The partial fraction of frac{3x - 1}{(1 - x + | vedclass

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Question

(C) Let $\frac{3x - 1}{(x^2 - x + 1)(x + 2)} = \frac{Ax + B}{x^2 - x + 1} + \frac{C}{x + 2}$.Multiplying both sides by $(x^2 - x + 1)(x + 2)$,we get:$

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$\frac{x}{x^2 - x + 1} - \frac{1}{x + 2}$

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(C) Let $\frac{3x - 1}{(x^2 - x + 1)(x + 2)} = \frac{Ax + B}{x^2 - x + 1} + \frac{C}{x + 2}$.Multiplying both sides by $(x^2 - x + 1)(x + 2)$,we get:$3x - 1 = (Ax + B)(x + 2) + C(x^2 - x + 1)$.To find $C$,put $x = -2$:$3(-2) - 1 = C((-2)^2 - (-2) + 1) \implies -7 = C(4 + 2 + 1) \implies -7 = 7C \implies C = -1$.Equating the coefficients of $x^2$:$0 = A + C \implies A = -C = 1$.Equating the constant terms:$-1 = 2B + C \implies -1 = 2B - 1 \implies 2B = 0 \implies B = 0$.Substituting the values of $A, B, C$:$\frac{3x - 1}{(x^2 - x + 1)(x + 2)} = \frac{x}{x^2 - x + 1} - \frac{1}{x + 2}$.

The partial fraction of $\frac{3x - 1}{(1 - x + x^2)(2 + x)}$ is:

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Given $\frac{3x-2}{(x+1)^2(x+3)} = \frac{A}{x+1} + \frac{B}{(x+1)^2} + \frac{C}{x+3}$,find the value of $4A + 2B + 4C$.

$\begin{aligned} & \frac{x^2+x+1}{(x-1)(x-2)(x-3)}=\frac{A}{x-1}+\frac{B}{x-2}+\frac{C}{x-3} \\ & \Rightarrow A+C= \end{aligned}$

Let $a, b$, and $c$ be such that $\frac{1}{(1-x)(1-2x)(1-3x)} = \frac{a}{1-x} + \frac{b}{1-2x} + \frac{c}{1-3x}$. Then $\frac{a}{1} + \frac{b}{3} + \frac{c}{5}$ is equal to:

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The coefficient of $x^3$ in the expansion of $\frac{x^4+1}{(x^2+1)(x-1)}$ when it is expressed in terms of positive integral powers of $x$,is

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